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Chapter 5: Problem 7

Give a recursive definition of the sequence \(\left\\{a_{n}\right\\}, n=\) \(1,2,3, \ldots\) if a) \(a_{n}=6 n\) b) \(a_{n}=2 n+1\) c) \(a_{n}=10^{n}\) d) \(a_{n}=5\)

Short Answer

Expert verified
Part (a): \(a_1=6\), \(a_n = a_{n-1} + 6\); Part (b): \(a_1 = 3\), \(a_n = a_{n-1} + 2\); Part (c): \(a_1 = 10\), \(a_n = 10 \times a_{n-1}\); Part (d): \(a_1 = 5\), \(a_n = 5\).

Step by step solution

01

Understanding Recursive Definition

A recursive definition provides a way to define the sequence by expressing terms using previous terms. It usually includes a base case and a recursive step.
02

Base Case - Part (a)

For the sequence \(a_n = 6n\), start with the first term. Thus, \(a_1 = 6 \times 1 = 6\).
03

Recursive Step - Part (a)

Assume \(a_{n}\) and \(a_{n-1}\) are terms in the sequence. We can express \(a_{n} \) as \(a_{n} = a_{n-1} + 6 \). Thus, the recursive formula is \(a_1 = 6 \) and \(a_n = a_{n-1} + 6 \).
04

Base Case - Part (b)

For the sequence \(a_n = 2n + 1\), start with the first term. Thus, \(a_1 = 2 \times 1 + 1 = 3\).
05

Recursive Step - Part (b)

Assume \(a_{n}\) and \(a_{n-1}\) are terms in the sequence. We can express \(a_{n}\) as \(a_{n} = a_{n-1} + 2 \). Thus, the recursive formula is \(a_1 = 3\) and \(a_n = a_{n-1} + 2 \).
06

Base Case - Part (c)

For the sequence \(a_n = 10^n\), start with the first term. Thus, \(a_1 = 10^1 = 10\).
07

Recursive Step - Part (c)

Assume \(a_{n}\) and \(a_{n-1}\) are terms in the sequence. We can express \(a_{n}\) as \( a_n = 10 \times a_{n-1} \). Thus, the recursive formula is \(a_1 = 10\) and \(a_n = 10 \times a_{n-1} \).
08

Base Case - Part (d)

For the sequence \(a_n = 5\), start with the first term. Thus, \(a_1 = 5\).
09

Recursive Step - Part (d)

No matter the value of \(n\), \(a_{n} = 5\). Hence, the recursive formula is \(a_1 = 5\) and \(a_n = a_{n-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Base Case
In recursive sequences, the base case is the initial, or 'starting', term of the sequence. It establishes the foundation from which all other terms are generated.

For instance, considering the sequence given by the problem, the base case is the value of the first term.

  • For the sequence \(a_n = 6n\): The base case is \(a_1 = 6\)
  • For the sequence \(a_n = 2n + 1\): The base case is \(a_1 = 3\)
  • For the sequence \(a_n = 10^n\): The base case is \(a_1 = 10\)
  • For the sequence \(a_n = 5\): The base case is \(a_1 = 5\)


The base case ensures that the sequence has a valued starting point which is essential for applying the recursive formula.
Recursive Formula Demystified
The recursive formula is another fundamental aspect in defining recursive sequences. This formula explains how each term in the sequence relates to the previous term or terms.

The term is defined by the previous term, which means you wouldn't explicitly define each term separately. Instead, you define the term using a pattern or rule that involves earlier terms in the sequence.

  • For \(a_n = 6n\), the recursive formula is \(a_1 = 6\) and \(a_n = a_{n-1} + 6\).
  • For \(a_n = 2n + 1\), the recursive formula is \(a_1 = 3\) and \(a_n = a_{n-1} + 2\).
  • For \(a_n = 10^n\), the recursive formula is \(a_1 = 10\) and \(a_n = 10 \times a_{n-1}\).
  • For \(a_n = 5\), the recursive formula is \(a_1 = 5\) and \(a_n = a_{n-1}\).


The recursive formula allows you to construct the entire sequence by repeatedly applying the same rule, ensuring consistency in the definition.
Sequence Termination Clarified
Sequence termination refers to the conditions under which a sequence ends or stabilizes.

In recursive sequences, some may continue indefinitely while others may reach a fixed point where they no longer change.

Consider the example sequences:
  • For \(a_n = 6n\): This sequence will grow indefinitely as each term increases by 6.
  • For \(a_n = 2n + 1\): This also continues indefinitely, increasing by 2 each step.
  • For \(a_n = 10^n\): This grows exponentially without termination.
  • For \(a_n = 5\): Despite the different values of \(n\), the sequence always remains at 5. This sequence stabilizes immediately.


Understanding whether a sequence terminates or stabilizes is key to analyzing its behavior over time and how it can be applied in various contexts.
Role of Discrete Mathematics
Discrete mathematics plays an essential role in understanding recursive sequences. Discrete mathematics deals with distinct and separate values, which is the nature of sequence terms.

Here, elements have a distinct difference from one another, making it easy to apply recursive definitions.

Recursive definitions are tools within discrete mathematics that help break down complex problems into simpler, manageable parts.

  • For instance, solving \(a_n = 6n\) involves discrete steps incrementally adding 6.
  • For \(a_n = 2n + 1\), each term is incremented by 2 following a linear progression.
  • \(a_n = 10^n\) uses exponential growth, another topic within discrete mathematics.
  • While \(a_n = 5\) represents a constant sequence, easily analyzed using discrete structures.


Recursive sequences in discrete mathematics help in developing algorithms, analyzing computer science problems, and solving mathematical puzzles efficiently.

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Most popular questions from this chapter

A partition of a positive integer \(n\) is a way to write \(n\) as a sum of positive integers where the order of terms in the sum does not matter. For instance, \(7=3+2+1+1\) is a partition of \(7 .\) Let \(P_{m}\) equal the number of different partitions of \(m,\) and let \(P_{m, n}\) be the number of different ways to express \(m\) as the sum of positive integers not exceeding \(n\). a) Show that \(P_{m, m}=P_{m}\) . b) Show that the following recursive definition for \(P_{m, n}\) is correct: $$P_{m, n}=\left\\{\begin{array}{ll}{1} & {\text { if } m=1} \\ {1} & {\text { if } n=1} \\ {P_{m, m}} & {\text { if } m< n} \\ {1+P_{m, m-1}} & {\text { if } m=n>1} \\ {P_{m, n-1}+P_{m-n, n}} & {\text { if } m>n>1}\end{array}\right.$$ c) Find the number of partitions of 5 and of 6 using this recursive definition.

Deal with iterations of the logarithm function. Let log \(n\) denote the logarithm of \(n\) to the base \(2,\) as usual. The function \(\log ^{(k)} n\) is defined recursively by $$\log ^{(k)} n=\left\\{\begin{array}{cl}{n} & {\text { if } k=0} \\ {\log \left(\log ^{(k-1)} n\right)} & {\text { if } \log ^{(k-1)} n \text { is defined }} \\ {} & {\text { and positive }} \\ {\text { undefined }} & {\text { otherwise. }}\end{array}\right.$$ The iterated logarithm is the function log" \(n\) whose value at \(n\) is the smallest nonnegative integer \(k\) such that \(\log ^{(k)} n \leq 1\) Find the value of \(\log ^{*} n\) for these values of \(n\) a) 2 b) 4 c) 8 d) 16 e) 256 f) 65536 g) \(2^{2048}\)

Use strong induction to show that if you can run one mile or two miles, and if you can always run two more miles once you have run a specified number of miles, then you can run any number of miles.

Pick's theorem says that the area of a simple polygon \(P\) in the plane with vertices that are all lattice points (that is, points with integer coordinates) equals \(I(P)+B(P) / 2-1\) where \(I(P)\) and \(B(P)\) are the number of lattice points in the interior of \(P\) and on the boundary of \(P,\) respectively. Use strong induction on the number of vertices of \(P\) to prove Pick's theorem. [Hint: For the basis step, first prove the theorem for rectangles, then for right triangles, and finally for all triangles by noting that the area of a tri- angle is the area of a larger rectangle containing it with the areas of at most three triangles subtracted. For the inductive step, take advantage of Lemma \(1 . ]\)

Find the reversal of the following bit strings. a) 0101 b) 11011 c) 100010010111
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